Answer:
c:(0;+∞)
Step-by-step explanation:
Given the situation, that the square root function only allows you values above than "0" (not equal neither), then you must consider that every value above 0 belongs to it's domain.
Then, to express the domain, going from your most negative number, to your most possitive number (in this case all positive number, thats why we use infinite) you must use the parenthesis wich means, you are not considering the value (in this case 0), but the value right after it, to the next value that as we said before, is inifinite. Also remember, that when you express a domain, and use infinite (despite it's going to negative way, or possitive way, it also goes with parenthesis).
<h3>
Answer: -10 and -40</h3>
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Explanation:
a = 200 = first term
d = -30 = common difference
Tn = nth term
Tn = a + d(n-1)
Tn = 200 + (-30)(n-1)
Tn = 200 - 30n + 30
Tn = -30n + 230
Set Tn less than 0 and isolate n
Tn < 0
-30n + 230 < 0
230 < 30n
30n > 230
n > 230/30
n > 7.667 approximately
Rounding up to the nearest whole number gets us 
So Tn starts to turn negative when n = 8
We can see that,
Tn = -30n + 230
T7 = -30*7 + 230
T7 = 20
and
Tn = -30n + 230
T8 = -30*8 + 230
T8 = -10 is the 8th term
and lastly
Tn = -30n + 230
T9 = -30*9 + 230
T9 = -40 is the ninth term
Or once you determine that T7 = 20, you subtract 30 from it to get 20-30 = -10 which is the value of T8. Then T9 = -40 because -10-30 = -40.
Answer:
7 is the (constant) coefficient of 1/y, but no, it is not the coefficient of y
Step-by-step explanation:
7/y can be rewritten as 7(1/y) or as 7(y^[-1]). In this case, yes, 7 is the (constant) coefficient of 1/y, but no, it is not the coefficient of y.
Answer:
Claim 2
Step-by-step explanation:
The Inscribed Angle Theorem* tells you ...
... ∠RPQ = 1/2·∠ROQ
The multiplication property of equality tells you that multiplying both sides of this equation by 2 does not change the equality relationship.
... 2·∠RPQ = ∠ROQ
The symmetric property of equality says you can rearrange this to ...
... ∠ROQ = 2·∠RPQ . . . . the measure of ∠ROQ is twice the measure of ∠RPQ
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* You can prove the Inscribed Angle Theorem by drawing diameter POX and considering the relationship of angles XOQ and OPQ. The same consideration should be applied to angles XOR and OPR. In each case, you find the former is twice the latter, so the sum of angles XOR and XOQ will be twice the sum of angles OPR and OPQ. That is, angle ROQ is twice angle RPQ.
You can get to the required relationship by considering the sum of angles in a triangle and the sum of linear angles. As a shortcut, you can use the fact that an external angle is the sum of opposite internal angles of a triangle. Of course, triangles OPQ and OPR are both isosceles.
